帮我看看这几题怎么做

2个回答

  • 一、1、lim[(sinx-sina)/(x-a)] (x->a)

    =lim[cosx/1] 洛必达法则

    =cosa

    2、lim(1+2x)^(1/x) (x->+∞)

    =lime^ln[(1+2x)^(1/x)]

    =e^limln[(1+2x)^(1/x)]

    =e^lim1/x*ln[(1+2x)]

    =e^lim2/(1+2x)/1 洛必达法则

    =e^0

    =1

    3、lim(1/x²-1/sin²x) (x->0)

    =lim[(sin²x-x²)/(x²sin²x)]

    =lim[(2sinxcosx-2x)/(2xsin²x+2x²sinxcosx)]

    =lim[(sin2x-2x)/(2xsin²x+x²sin2x)]

    =lim[(2cos2x-2)/(2sin²x+4xsinxcosx+2xsin2x+2x²cos2x)]

    =lim[(cos2x-1)/(sin²x+2xsin2x+x²cos2x)]

    =lim[(-2sin2x)/(2sinxcosx+2sin2x+4xcos2x+2xcos2x-2x²sin2x)]

    =lim[(-2sin2x)/(3sin2x+6xcos2x-2x²sin2x)]

    =lim[(-4cos2x)/(6cos2x+6cos2x-12xsin2x-4xsin2x-4x²cos2x)]

    =(-4)/(6+6+0)

    =-1/3

    就是不断的用洛必达法则求导

    4、1+1/2+1/3+1/4+...1/n 为调和级数,

    其和为 ln(n+1) + r,r为欧拉常数,n->∞

    ∴lim(1+1/2+1/3+1/4+...1/n)^(1/n)

    =lim[ln(n+1) + r]^(1/n)

    =e^lim[ln(n+1)+r]/n

    =e^lim[1/(n+1)]/1

    =e^0

    =1

    5、y=[1+e^(-x^2)]/[1-e^(-x^2)]

    x->0时,limy=(1+1)/(1-1)=+∞

    x->+∞时,limy=(1+0)/(1-0)=1

    ∴该函数有垂直渐近线x=0,水平渐近线y=1

    二、1、f(x)=(e^x+log3(x))arcsinx

    f'(x)=(e^x+log3(x))'arcsinx+(e^x+log3(x))*(arcsinx)'

    =[(e^x)'+(log3(x))']arcsinx+(e^x+log3(x))*1/√(1-x^2)

    =[e^x+1/(xln3)]arcsinx+(e^x+log3(x))*1/√(1-x^2)

    2、x=t^2sint,y=t^2cost

    dx/dt=2tsint+t^2cost

    dy/dt=2tcost-t^2sint

    dy/dx=(dy/dt)/(dx/dt)

    =(2tsint+t^2cost)/(2tcost-t^2sint)

    =(2sint+tcost)/(2cost-tsint)